Convergent Real Sequence/Examples/Arithmetic Mean of Previous 2 Terms

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Example of Convergent Real Sequence

Let $\sequence {x_n}_{n \mathop \in \N_{>0} }$ be the sequence in $\R$ defined as:

$x_n = \begin {cases} a & : n = 1 \\ b & : n = 2 \\ \dfrac {x_{n - 1} + x_{n - 2} } 2 & : n > 2 \end {cases}$

That is, beyond the first $2$ terms, each term is the arithmetic mean of the previous $2$ terms.

Then $\sequence {x_n}$ converges.


Proof

\(\displaystyle x_{n + 2} - x_{n + 1}\) \(=\) \(\displaystyle \dfrac {x_{n + 1} + x_n} 2 - x_{n + 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {x_n - x_{n + 1} } 2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \size {x_{n + 2} - x_{n + 1} }\) \(=\) \(\displaystyle \dfrac {\size {x_n - x_{n + 1} } } 2\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\size {x_n - x_{n - 1} } } {2^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \ldots\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\size {x_2 - x_1} } {2^n}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\size {b - a} } {2^n}\)


Let $n > m$.

Then:

\(\displaystyle \size {x_n - x_m}\) \(=\) \(\displaystyle \size {\paren {x_n - x_{n - 1} } + \paren {x_{n - 1} - x_{n - 2} } + \dotsb + \paren {x_{m + 1} - x_m} }\)
\(\displaystyle \) \(\le\) \(\displaystyle \size {x_n - x_{n - 1} } + \size {x_{n - 1} - x_{n - 2} } + \dotsb + \size {x_{m + 1} - x_m}\) Triangle Inequality for Real Numbers
\(\displaystyle \) \(\le\) \(\displaystyle \paren {\dfrac 1 {2^{n - 2} } + \dfrac 1 {2^{n - 3} } + \dotsb + \dfrac 1 {2^{m - 1} } } \size {b - a}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 {2^{m - 1} } \paren {1 + \dfrac 1 2 + \dfrac 1 {2^2} + \dotsb + \dfrac 1 {2^{n - m - 1} } } \size {b - a}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 {2^{m - 1} } \paren {\dfrac {1 - \paren {\frac 1 2}^{n - m} } {1 - \frac 1 2} } \size {b - a}\) Sum of Geometric Progression
\(\displaystyle \) \(\le\) \(\displaystyle \dfrac 1 {2^{m - 2} } \size {b - a}\)


Let $\epsilon \in \R_{>0}$ be given.

Let $N$ be sufficiently large that:

$\dfrac 1 {2^{N - 2} } \size {b - a} < \epsilon$


Then:

$\forall n > N, m > N: \size {x_n - x_m} \le \dfrac 1 {2^{N - 2} } \size {b - a} < \epsilon$

Hence it is seen that $\sequence {x_n}$ is a Cauchy sequence.

Hence the result, by Cauchy Sequence Converges on Real Number Line.

$\blacksquare$


Sources